What theorem is being invoked in this solution to deciding if this polynomial is solvable by radicals?

Question

I was reading these solutions online http://campus.lakeforest.edu/trevino/Spring2019/Math331/Homework7Solutions.pdf for practice for an upcoming exam I have . But in question $2$ part $(a)$ the author uses a trick I'm not familiar with , he takes the derivative of the polynomial $x^5-12x^2+2$ , notes that the roots of said derivative are $0$ and $\sqrt[3]{\tfrac{24}{5}}$ and then from this deduces that there are at most 3 real roots to our original polynomial .

What theorem is the author invoking here ?

Answer 1

Just to expand on my comment, and to see that no other theorem (certainly not IVT or the Fundamental Theorem) need to be used here:

The standard version of Rolle's Theorem asserts that if a real-valued function is continuous on the interval $[a,b]$ and differentiable inside that interval, and $f(a) = f(b)$, then there is at least one $c \in (a,b)$ with $f'(c)=0$.

Now, since any polynomial $p$ is differentiable and hence continuous everywhere, Rolle's Theorem asserts that between any two of its zeroes (taking the role of $a$ and $b$ above), its derivative $p'$ must also have a zero. This easily generalises to:

If $p$ has $n$ different zeroes, then $p'$ has at least $n-1$ different zeroes

(namely, at least one zero of $p'$ between the first and second zero of $p$, then at least one between the second and third, ..., and at least one between the $n-1$th and the $n$th zero of $p$, where by "first", "second" etc. I just order the zeroes from left to right on the real number line).

Which, when you think about it for a moment, implies that:

If $p'$ has $n-1$ different zeroes, then $p$ cannot have more than $n$ different zeroes

which is what is used in your case, where $n=3$.